Newton’s law of cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. So the rate of cooling for a bottle of lemonade at a room temperature of 75°F which is placed into a refrigerator with temperature of 38°F can be modeled by dT dt equals k times the quantity T minus 38 where T(t) is the temperature of the lemonade after t minutes and T(0) = 75. After 30 minutes the lemonade has cooled to 60°F, so T(30) = 60.
To the nearest degree, what is the temperature of the lemonade after an additional 30 minutes?
To the nearest minute, how long does it take for the lemonade to cool to 55°F?
To find the temperature of the lemonade after an additional 30 minutes, we need to solve the differential equation:
dT/dt = k(T - 38)
using the given initial condition T(0) = 75.
To solve this differential equation, we can separate variables and integrate both sides. This gives:
∫ dT / (T - 38) = ∫ k dt
To solve the integral, we can use a logarithmic substitution. Let u = T - 38, then dT = du. Substituting these into the integral, we have:
∫ du / u = ∫ k dt
ln|u| = kt + C1
Using the initial condition T(0) = 75, we have T(0) - 38 = 75 - 38 = 37, so u(0) = 37. Substituting this into the equation above, we get:
ln|37| = C1
Simplifying further, we have:
ln(37) = C1
Now we can substitute back to solve for T(t). Rearranging the equation, we have:
ln|T - 38| = kt + ln(37)
To find the temperature of the lemonade after an additional 30 minutes, we can set t = 30 and solve for T:
ln|T(30) - 38| = k(30) + ln(37)
Since T(30) = 60 (given), we have:
ln|60 - 38| = k(30) + ln(37)
ln|22| = 30k + ln(37)
Using logarithmic property, we have:
22 = e^(30k + ln(37))
Now we can solve for k. Taking the natural logarithm of both sides, we get:
ln(22) = 30k + ln(37)
30k = ln(22) - ln(37)
k = (ln(22) - ln(37))/30
Plugging this value of k into the equation above, we can solve for T(60):
ln|T(60) - 38| = [(ln(22) - ln(37))/30](60) + ln(37)
Simplifying this equation will give us the temperature of the lemonade after an additional 30 minutes.
To find how long it takes for the lemonade to cool to 55°F, we can set T = 55 and solve for t. Rearranging the equation, we have:
ln|55 - 38| = kt + ln(37)
ln|17| = kt + ln(37)
Using the value of k from above, we can solve for t:
ln(17) = [(ln(22) - ln(37))/30]t + ln(37)
Simplifying this equation will give us the time it takes for the lemonade to cool to 55°F.
To find the temperature of the lemonade after an additional 30 minutes, we can use the given information and the differential equation representing the rate of cooling.
The differential equation is given as:
dT/dt = k * (T - 38)
From the given information, we know that T(0) = 75 and T(30) = 60.
To find the value of k, we can use the fact that T(30) = 60. We substitute these values into the equation and solve for k:
60 = k * (75 - 38)
60 = k * 37
k = 60 / 37
Now that we have the value of k, we can solve the differential equation to find T(60).
dT/dt = k * (T - 38)
Separate the variables by multiplying both sides by dt:
dT / (T - 38) = k * dt
Integrate both sides:
∫ dT / (T - 38) = k ∫ dt
ln|T - 38| = k * t + C
where C is the constant of integration.
Now we can find T(60) using the initial condition T(0) = 75:
ln|T(60) - 38| = (60 / 37) * 60 + C1
Substitute T(0) = 75:
ln|75 - 38| = (60 / 37) * 0 + C1
ln|37| = C1
Therefore, the equation becomes:
ln|T(60) - 38| = (60 / 37) * 60 + ln|37|
Now we can find the temperature T(60) by solving for T:
|T(60) - 38| = e^((60 / 37) * 60 + ln|37|)
Solving for T(60):
T(60) - 38 = e^((60 / 37) * 60 + ln|37|)
T(60) = 38 + e^((60 / 37) * 60 + ln|37|)
Now we can substitute the given values and calculate the temperature to the nearest degree:
T(60) = 38 + e^((60 / 37) * 60 + ln|37|)
T(60) ≈ 46.27°F (rounded to the nearest degree)
To find how long it takes for the lemonade to cool to 55°F, we can follow a similar process. We need to solve the equation for T(t) = 55:
ln|T(t) - 38| = (60 / 37) * t + ln|37|
|T(t) - 38| = e^((60 / 37) * t + ln|37|)
Solving for t:
log((T(t) - 38)) = (60 / 37) * t + ln|37|
t = (1 / (60 / 37)) * [log((T(t) - 38)) - ln|37|]
Substituting T(t) = 55:
t = (1 / (60 / 37)) * [log((55 - 38)) - ln|37|]
t ≈ 48.2 minutes (rounded to the nearest minute)
Therefore, it takes approximately 48 minutes for the lemonade to cool to 55°F (rounded to the nearest minute).
dT/dt = -k (T-38)
try
dT/(T-38) = -k dt
ln(T-38) = -k t
T-38 = ce^-kt
T = 38 + ce^-kt
when t = 0, T =75
75 = 38 + c
c = 37
so T = 38 + 37 e^-kt
at t = 30 , T = 60
so
60 = 38 + 37 e^-30 k
22/37 = .595 = e^-30 k
ln .595 = -.52 = -30 k
so k = .0173
and
T = 38 + 37 e^-.0173 t
you can take it from there.